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It is surely at the low end of what is interesting here. I recommend AoPS.
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Greg KuperbergNov 6 '09 at 14:34

2

Upvoted because, although the question isn't really appropriate, the answers are great.
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David SpeyerNov 6 '09 at 16:14

Another way of seeing this is to compare two consecutive square numbers, say n2 and (n+1)2. If we expand the larger one, we get (n+1)2=n2+2n+1, so it is exactly (2n+1) more than the previous square. As n increases, this expression represents the consecutive odd numbers. So squares always differ by consecutive odd numbers.

I know you didn't ask for a proof, but actually the best way of seeing WHY something happens is understanding a good, simple proof for it. Here is my try to your nice observation, using the formula for the sum of several consecutive terms of an arithmetic progression (which, by the way, has also a nice, simple proof):

You know that if $\{an+b\}$ is an arithmetic progression and you look at some of its consecutive terms, then their sum is "(the first one plus the last one) times (the number of terms added) divided by 2".

In your case, you have the progression $\{2n-1\}$, starting at $1$ and finishing in some $2N-1$, which has $N$ terms. Then, by the formula above, you get that their sum is $$S = \frac{(1+2N-1)\cdot N}{2} = \frac{2N\cdot N}{2} = N^2.$$

This, if you look at it by the reverse side, is caused by the fact that the substraction of two consecutive squares is a particular odd number, $(N+1)^2 - N^2 = N^2+2N+1-N^2 = 2N+1$. You could also use this to prove your claim by induction, but it wouldn't be, in my opinion, a clarifying proof of the kind you were looking for.